Quick Analysis Problem - Related to properties of cont functions

brickcitie
Messages
15
Reaction score
0
Quick Analysis Problem -- Related to properties of cont functions

Homework Statement



This problem assumes that f: R ----> R is continuous on all of R. I need to determine if the following set is guaranteed to be closed, regardless of f(x).

A = {x in R | 0 <= f(x) <= 1}



Homework Equations



Everything is contained above.

The Attempt at a Solution



Ok so I have tried for a while to come up with a counter example. My first idea was to let f(x) = 1/ x^2 + 1, but then A = R which is closed and doesn't help my case. My next idea was to let f(x) = sinx, but then A is an infinite union of closed intervals on R, which I believe is a closed set.

I cannot think of any counter-example, so I'm guessing A will be guaranteed to be closed, but I cannot really get a formal proof started. Any idea?
 
Last edited:
Physics news on Phys.org


Take f(x)=x. Then A is not open...


Also note that an infinite union of closed sets is not necessairly closed. You mentioned that, but it is incorrect.
 


Hi again micromass. So I mis-typed this originally; I meant to say that I need to decide whether it is guaranteed to be closed. The first part of the question was whether it was open or not, and I used the same function you just mentioned to answer that.

And also, won't any infinite union of closed disjoint intervals in R be closed, because the complement will be an infinite union of open intervals, and thus open?
 


Oh, I see.

Well do you know that a function is continuous if and only if for every open set G holds that f^{-1}(G) is open??

As for the second part. The complement of an infinite union of closed sets is an infinite intersection of open sets, and this is not guaranteed to be open.
But if you take the infinite union of DISJOINT closed sets, then it is indeed closed. But not for the reason you stated...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top